Question

ABCD is a trapezium in which ab parallel to dc a b is equal to 7 cm is equal to BC is equal to 5 cm and distance between AC and DC is 4 cm find the length of BC and hence find area of trapezium ABCD

Answer 1

Area of trapezium ABCD is $180cm^2$  and the length of BC is 13 cm.

Step-by-step explanation:

Given:

Her we have, AB = 7 cm, AD = BC = 5 cm

Height = AL = BM = 4cm  

Here in the given figure, AB = LM

LM = 7 cm                    [1]

Now, consider  ΔALD ,

By Pythagoras theorem , we have

$AD^2 = AL^2 + DL^2$

$5^2 = 4^2 + DL^2$

$DL^2 = 5^2 - 4^2 = 25 - 16 = 9$

DL = 3 cm                 [2]

Similarly in  Δ BMC ,

By Pythagoras theorem

$BC^2 = BM^2 + MC^2$

$5^2 = 4^2 + MC^2$

$MC^2 = 5^2 - 4^2 = 25 - 16 = 9$

MC = 3 cm               [3]

From Eq (1), (2) and (3 ), we get

DC = DL + LM + MC = 3 + 7 + 3 = 13 cm

We know that,

Area of trapezium = $\frac{1}{2}\times (sum of parallel sides)\times height$

Area of trapezium = $\frac{1}{2}\times (AB + DC) \times AL$

                                = $\frac{1}{2}\times (7 + 13) \times 4 = 40 cm^2$

Therefore, area of trapezium ABCD is $180cm^2$ .